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%I #21 Sep 07 2018 16:44:41
%S 1,1,1,1,3,3,1,9,12,16,1,25,60,80,125,1,75,330,480,750,1296,1,231,
%T 1680,3920,5250,9072,16807,1,763,9408,33600,49000,72576,134456,262144,
%U 1,2619,56952,254016,598500,762048,1210104,2359296,4782969,1,9495,348120
%N Triangle read by rows: T(n,k) = number of forests on n labeled nodes, where k is the maximum of the number of edges per tree (n>=1, 0<=k<=n-1).
%H Alois P. Heinz, <a href="/A143911/b143911.txt">Rows n = 1..141, flattened</a>
%F See program.
%e T(4,1) = 9, because 9 forests on 4 labeled nodes have 1 as the maximum of the number of edges per tree:
%e .1-2. .1.2. .1.2. .1.2. .1.2. .1.2. .1-2. .1.2. .1.2.
%e ..... ...|. ..... .|... ..\.. ../.. ..... .|.|. ..X..
%e .4.3. .4.3. .4-3. .4.3. .4.3. .4.3. .4-3. .4.3. .4.3.
%e Triangle begins:
%e 1;
%e 1, 1;
%e 1, 3, 3;
%e 1, 9, 12, 16;
%e 1, 25, 60, 80, 125;
%e 1, 75, 330, 480, 750, 1296;
%p A:= (n,k)-> coeff(series(exp(add(j^(j-2) *x^j/j!, j=1..k)), x, n+1), x,n)*n!: T:= (n,k)-> A(n,k+1)-A(n,k): seq(seq(T(n,k), k=0..n-1), n=1..11);
%t A[n_, k_] := SeriesCoefficient[Exp[Sum[j^(j-2)*x^j/j!, {j, 1, k}]], {x, 0, n}]*n!; T[n_, k_] := A[n, k+1] - A[n, k];
%t Table[T[n, k], {n, 1, 11}, {k, 0, n-1}] // Flatten (* _Jean-François Alcover_, May 31 2016, translated from Maple *)
%Y Columns k=0-1 give: A000012, A001189.
%Y Row sums give A001858.
%Y Rightmost diagonal gives A000272.
%Y Cf. A138464.
%K nonn,tabl
%O 1,5
%A _Alois P. Heinz_, Sep 04 2008
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